<span>1 of them i think try it out</span>
Let's say the Moon's complete cycle of phases is about 28 days (4 weeks) ... just so the arithmetic is easier. (Close enough. It's actually 29.53 days.)
If it's halfway through the waxing crescent, then it'll reach first quarter in 3 days or so.
From there, it'll take the next 7 days to become Full, and another 7 days to reach last quarter.
Total from half-waxing-crescent to last quarter . . . about 17 days.
<span>θ is the angle whose tangent is 3/0.4 = arctan (7.5) = 82.4°
(That's the angle above the horizontal. If that doesn't match
the </span><span>θ in your diagram, too bad ! For some strange reason,
I wasn't able to see your diagram.)
</span>
Answer:
1.324 × 10⁷ m
Explanation:
The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.
Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.
We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have
Rω² = GME/R²
R(2π/T)² = GME/R²
R³ = GME(T/2π)²
R = ∛(GME)(T/2π)²
RE + h = ∛(GMET²/4π²)
h = ∛(GMET²/4π²) - RE
substituting the values of the variables, we have
h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m
h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m
h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m
h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m
h = 19.64 × 10⁶ m - 6.4 × 10⁶ m
h = 13.24 × 10⁶ m
h = 1.324 × 10⁷ m
Answer:
Explanation:
Since the force applied is parallel to the displacement of the car, the work done on the car is simply given by:
where
F = 1210 N is the force applied on the car
d = 201 m is the displacement of the car
Substituting numbers into the equation, we find:
